For the galvanic cell,Zn_{(s)} + Cu^{2+}(0.02 | vedclass

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Question

(B) The cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$Standard cell potential $E_{cell}^{0} = E_{cathode}^{0} - E

Vedclass · Accepted Answer

$109$

Vedclass · Answer

(B) The cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} ightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$Standard cell potential $E_{cell}^{0} = E_{cathode}^{0} - E_{anode}^{0} = 0.34 \ V - (-0.76 \ V) = 1.10 \ V$Using the Nernst equation at $298 \ K$:$E_{cell} = E_{cell}^{0} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,$[Zn^{2+}] = 0.04 \ M$,and $[Cu^{2+}] = 0.02 \ M$$E_{cell} = 1.10 - \frac{0.059}{2} \log \frac{0.04}{0.02}$$E_{cell} = 1.10 - 0.0295 	imes \log(2)$$E_{cell} = 1.10 - 0.0295 	imes 0.3010 \approx 1.10 - 0.00888 = 1.09112 \ V$$E_{cell} = 109.112 	imes 10^{-2} \ V$Rounding to the nearest integer,we get $109$.

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

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For the galvanic cell,$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$$[\frac{2.303 \ RT}{F} = 0.059 \ V]$